Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x-4y &= -3 \\ 6x+9y &= -2\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $6x = -9y-2$ Divide both sides by $6$ to isolate $x$ $x = {-\dfrac{3}{2}y - \dfrac{1}{3}}$ Substitute this expression for $x$ in the first equation. $2({-\dfrac{3}{2}y - \dfrac{1}{3}}) - 4y = -3$ $-3y - \dfrac{2}{3} - 4y = -3$ Simplify by combining terms, then solve for $y$ $-7y - \dfrac{2}{3} = -3$ $-7y = -\dfrac{7}{3}$ $y = \dfrac{1}{3}$ Substitute $\dfrac{1}{3}$ for $y$ in the top equation. $2x-4( \dfrac{1}{3}) = -3$ $2x-\dfrac{4}{3} = -3$ $2x = -\dfrac{5}{3}$ $x = -\dfrac{5}{6}$ The solution is $\enspace x = -\dfrac{5}{6}, \enspace y = \dfrac{1}{3}$.